Question: $f\,^{\prime}(x)=\dfrac{100}{x^3}$ and $f(5)=-14$. $f(1) = $
Explanation: Finding $f(x)$ We have $f'(x)=\dfrac{100}{x^3}$ and we want to find $f(x)$ : $\begin{aligned}f(x)& = \int f'(x)\,dx \\\\ & = \int \dfrac{100}{x^3}\,dx \\\\ & = {-50x^{-2}} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(5)=-14$. Here's what we get when we plug in $5$ : $\begin{aligned}f(5)&={-50(5)^{-2}} {+ C}\\\\ &={-2} {+ C} \end{aligned}$ We are given that this must equal $-14$ : $-14 = {-2} {+ C}$ Solving the equation gives us ${C=-12}$. Finding $f(1)$ Now, we have that $f(x)= {50x^{-2}} {-12}$. Let's find $f(1)$ by plugging in $1$ : $\begin{aligned}f(1)&=-50(1)^{-2}-12\\\\ &=-62 \end{aligned}$ The answer $f(1) =- 62$